Chapter 23, Problem 12P

(a)

To determine

The magnitude and direction of the net electric force on $q_1$.

Answer to Problem 12P

The direction of net electric force is along negative $x$ axis and the magnitude of net electric force on $q_1$ is $-133.052\text{N}$.

Explanation of Solution

Write the expression for electrostatic force.

    $F=k_{\text{e}}\frac{q_1{q}_2}{r^2}$

Here, $F$ is the electrostatic force, $k_{\text{e}}$ is a coulomb’s constant, $r$ is the distance between two charges, $q_1$ and $q_2$ are the first and second point charges respectively.

The force acting on three charges is shown in Figure (1).

Figure (1)

Write the expression for electric force due to $q_2$ charge.

    ${\overrightarrow{F}}_{21}=-k_{\text{e}}\frac{q_2{q}_1}{d_1^2}\widehat{\text{i}}$

Here, $d_1$ is the distance between $q_1$ and $q_2$, ${\overrightarrow{F}}_{21}$ is the electric force due to $q_2$.

Write the expression for electric force due to $q_3$ charge.

    ${\overrightarrow{F}}_{31}=k_{\text{e}}\frac{q_3{q}_1}{{\left(d_1+d_2\right)}^2}\widehat{\text{i}}$

Here, $d_2$ is the distance between $q_2$ and $q_3$, ${\overrightarrow{F}}_{31}$ is the electric force due to $q_3$.

Write the expression for electric force on $q_1$ charge due to $q_2$ and $q_3$ charges.

    ${\overrightarrow{F}}_1={\overrightarrow{F}}_{21}+{\overrightarrow{F}}_{31}$

Substitute $-k_{\text{e}}\frac{q_2{q}_1}{d_1^2}\widehat{\text{i}}$ for ${\overrightarrow{F}}_{21}$ and $k_{\text{e}}\frac{q_3{q}_1}{{\left(d_1+d_2\right)}^2}\widehat{\text{i}}$ for ${\overrightarrow{F}}_{31}$ in the above equation to find ${\overrightarrow{F}}_1$.

    ${\overrightarrow{F}}_1=-k_{\text{e}}\frac{q_2{q}_1}{d_1^2}\widehat{\text{i}}+k_{\text{e}}\frac{q_3{q}_1}{{\left(d_1+d_2\right)}^2}\widehat{\text{i}}$                                                                   (I)

Conclusion:

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_{\text{e}}$, $6.00\mu \text{C}$ for $q_1$, $-2.00\mu \text{C}$ for $q_3$, $1.50\mu \text{C}$ for $q_2$, $2.00\text{cm}$ for $d_2$ and $3.00\text{cm}$ for $d_1$ in equation (I) to find ${\overrightarrow{F}}_1$.

    $\begin{array}{c}{\overrightarrow{F}}_1=-\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\frac{\left[\left(1.50\mu \text{C}\right)\left(\frac{1\times {10}^{-6}\text{C}}{1\mu \text{C}}\right)\right]\left[\left(6.00\mu \text{C}\right)\left(\frac{1\times {10}^{-6}\text{C}}{1\mu \text{C}}\right)\right]}{{\left[\left(3.00\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}^2}\widehat{\text{i}}+\\ \left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\frac{\left[\left(-2.00\mu \text{C}\right)\left(\frac{1\times {10}^{-6}\text{C}}{1\mu \text{C}}\right)\right]\left[\left(6.00\mu \text{C}\right)\left(\frac{1\times {10}^{-6}\text{C}}{1\mu \text{C}}\right)\right]}{{\left[\left(5.00\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}^2}\widehat{\text{i}}\\ {\overrightarrow{F}}_1=\left(-89.9\text{N}-43.152\text{N}\right)\widehat{\text{i}}\\ =\left(-133.052\text{N}\right)\widehat{\text{i}}\end{array}$

Therefore, the direction of net electric force is along negative $x$ axis and the magnitude of net electric force on $q_1$ is $-133.052\text{N}$.

(b)

To determine

The magnitude and direction of the net electric force on $q_2$.

Answer to Problem 12P

The direction of net electric force is along positive $x$ axis and the magnitude of net electric force on $q_2$ is $22.475\text{N}$.

Explanation of Solution

Write the expression for electric force due to $q_1$ charge.

    ${\overrightarrow{F}}_{12}=k_{\text{e}}\frac{q_1{q}_2}{d_1^2}\widehat{\text{i}}$

Here, ${\overrightarrow{F}}_{12}$ is the electric force due to $q_1$.

Write the expression for electric force due to $q_3$ charge.

    ${\overrightarrow{F}}_{32}=k_{\text{e}}\frac{q_3{q}_2}{{\left(d_2\right)}^2}\widehat{\text{i}}$

Here, ${\overrightarrow{F}}_{32}$ is the electric force due to $q_3$.

Write the expression for electric force on $q_2$ charge due to $q_1$ and $q_3$ charges.

    ${\overrightarrow{F}}_2={\overrightarrow{F}}_{12}+{\overrightarrow{F}}_{32}$

Substitute $k_{\text{e}}\frac{q_1{q}_2}{d_1^2}\widehat{\text{i}}$ for ${\overrightarrow{F}}_{12}$ and $k_{\text{e}}\frac{q_3{q}_2}{{\left(d_2\right)}^2}\widehat{\text{i}}$ for ${\overrightarrow{F}}_{32}$ in the above equation to find ${\overrightarrow{F}}_2$.

    ${\overrightarrow{F}}_2=k_{\text{e}}\frac{q_1{q}_2}{d_1^2}\widehat{\text{i}}+k_{\text{e}}\frac{q_3{q}_2}{{\left(d_2\right)}^2}\widehat{\text{i}}$                                                                  (II)

Conclusion:

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_{\text{e}}$, $6.00\mu \text{C}$ for $q_1$, $-2.00\mu \text{C}$ for $q_3$, $1.50\mu \text{C}$ for $q_2$, $2.00\text{cm}$ for $d_2$ and $3.00\text{cm}$ for $d_1$ in equation (II) to find ${\overrightarrow{F}}_2$.

    $\begin{array}{c}{\overrightarrow{F}}_2=\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\frac{\left[\left(1.50\mu \text{C}\right)\left(\frac{1\times {10}^{-6}\text{C}}{1\mu \text{C}}\right)\right]\left[\left(6.00\mu \text{C}\right)\left(\frac{1\times {10}^{-6}\text{C}}{1\mu \text{C}}\right)\right]}{{\left[\left(3.00\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}^2}\widehat{\text{i}}+\\ \left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\frac{\left[\left(-2.00\mu \text{C}\right)\left(\frac{1\times {10}^{-6}\text{C}}{1\mu \text{C}}\right)\right]\left[\left(1.50\mu \text{C}\right)\left(\frac{1\times {10}^{-6}\text{C}}{1\mu \text{C}}\right)\right]}{{\left[\left(2.00\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}^2}\widehat{\text{i}}\\ {\overrightarrow{F}}_2=\left(89.9\text{N}-67.425\text{N}\right)\widehat{\text{i}}\\ =\left(22.475\text{N}\right)\widehat{\text{i}}\end{array}$

Therefore, the direction of net electric force is along positive $x$ axis and the magnitude of net electric force on $q_2$ is $22.475\text{N}$.

(c)

To determine

The magnitude and direction of the net electric force on $q_3$.

Answer to Problem 12P

The direction of net electric force is along negative $x$ axis and the magnitude of net electric force on $q_3$ is $110.5\text{N}$.

Explanation of Solution

Write the expression for electric force due to $q_1$ charge.

    ${\overrightarrow{F}}_{13}=-k_{\text{e}}\frac{q_1{q}_3}{{\left(d_1+d_2\right)}^2}\widehat{\text{i}}$

Here, ${\overrightarrow{F}}_{13}$ is the electric force due to $q_1$.

Write the expression for electric force due to $q_2$ charge.

    ${\overrightarrow{F}}_{23}=-k_{\text{e}}\frac{q_2{q}_3}{{\left(d_2\right)}^2}\widehat{\text{i}}$

Here, ${\overrightarrow{F}}_{23}$ is the electric force due to $q_2$.

Write the expression for electric force on $q_3$ charge due to $q_1$ and $q_2$ charges.

    ${\overrightarrow{F}}_3={\overrightarrow{F}}_{13}+{\overrightarrow{F}}_{23}$

Substitute $-k_{\text{e}}\frac{q_1{q}_3}{{\left(d_1+d_2\right)}^2}\widehat{\text{i}}$ for ${\overrightarrow{F}}_{13}$ and $-k_{\text{e}}\frac{q_2{q}_3}{{\left(d_2\right)}^2}\widehat{\text{i}}$ for ${\overrightarrow{F}}_{23}$ in the above equation to find ${\overrightarrow{F}}_3$.

    ${\overrightarrow{F}}_3=-k_{\text{e}}\frac{q_1{q}_3}{{\left(d_1+d_2\right)}^2}\widehat{\text{i}}+\left(-k_{\text{e}}\frac{q_2{q}_3}{{\left(d_2\right)}^2}\right)\widehat{\text{i}}$                                                                (III)

Conclusion:

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_{\text{e}}$, $6.00\mu \text{C}$ for $q_1$, $-2.00\mu \text{C}$ for $q_3$, $1.50\mu \text{C}$ for $q_2$, $2.00\text{cm}$ for $d_2$ and $3.00\text{cm}$ for $d_1$ in equation (III) to find ${\overrightarrow{F}}_3$.

    $\begin{array}{c}{\overrightarrow{F}}_3=-\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\frac{\left[\left(-2.00\mu \text{C}\right)\left(\frac{1\times {10}^{-6}\text{C}}{1\mu \text{C}}\right)\right]\left[\left(6.00\mu \text{C}\right)\left(\frac{1\times {10}^{-6}\text{C}}{1\mu \text{C}}\right)\right]}{{\left[\left(5.00\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}^2}\widehat{\text{i}}-\\ \left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\frac{\left[\left(-2.00\mu \text{C}\right)\left(\frac{1\times {10}^{-6}\text{C}}{1\mu \text{C}}\right)\right]\left[\left(1.50\mu \text{C}\right)\left(\frac{1\times {10}^{-6}\text{C}}{1\mu \text{C}}\right)\right]}{{\left[\left(2.00\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}^2}\widehat{\text{i}}\\ {\overrightarrow{F}}_3=\left(43.152\text{N}+67.425\text{N}\right)\widehat{\text{i}}\\ {\overrightarrow{F}}_3=\left(110.5\text{N}\right)\widehat{\text{i}}\end{array}$

Therefore, the direction of net electric force is along negative $x$ axis and the magnitude of net electric force on $q_3$ is $110.5\text{N}$.